A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semivertical angle is $$\tan^{-1}\frac{3}{4}$$. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is _______.

A right circular cone has its vertex downward and semivertical angle $$\theta = \tan^{-1}\frac{3}{4}$$. Water is poured into the cone at a constant rate of 6 cubic meters per hour, and we seek the rate of increase of the wet curved surface area when the depth of the water is $$h = 4$$ meters.

Denote by $$h the depth of the water and by r the radius of the water surface. Since \tan\theta = \frac{3}{4}$$ and the triangle formed by the radius and the depth is similar to the triangle defined by the semivertical angle, we have $$\frac{r}{h} = \frac{3}{4}, hence r = \frac{3h}{4}$$.

The volume of water in the cone is given by $$V = \frac{1}{3}\pi r^2 h, which on substituting for r becomes V = \frac{1}{3}\pi\left(\frac{3h}{4}\right)^2 h = \frac{3\pi h^3}{16}\,. Differentiating with respect to time t yields \frac{dV}{dt} = \frac{9\pi h^2}{16}\cdot\frac{dh}{dt}\,, and since \frac{dV}{dt} = 6, we have \frac{9\pi h^2}{16}\frac{dh}{dt} = 6\,. At h = 4 this gives 9\pi\frac{dh}{dt} = 6 and thus \frac{dh}{dt} = \frac{2}{3\pi}\,.$$

Next, the slant height $$l of the water surface satisfies l = \frac{h}{\cos\theta}. From \tan\theta = \frac{3}{4} one finds \sin\theta = \frac{3}{5} and \cos\theta = \frac{4}{5}, so l = \frac{5h}{4}\,. The wet curved surface area is S = \pi r\,l = \pi\cdot\frac{3h}{4}\cdot\frac{5h}{4} = \frac{15\pi h^2}{16}\,. $$

Differentiating $$S with respect to time gives \frac{dS}{dt} = \frac{15\pi\cdot 2h}{16}\cdot\frac{dh}{dt} = \frac{30\pi h}{16}\cdot\frac{dh}{dt}\,. Substituting h = 4 and \frac{dh}{dt} = \frac{2}{3\pi} yields \frac{dS}{dt} = \frac{30\pi\cdot 4}{16}\cdot\frac{2}{3\pi} = 5\,. $$ Therefore, the wet curved surface area is increasing at the rate $$5$$ square meters per hour.